Is aob = a^b is a binary operation on N? : Group Theory

 

Problem  :

  Show that the operation ‘ o ‘ given by aob = a^b is a binary operation on the set of natural numbers N. Is this operation associative and commutative in N?

Solution :

   Consider N,the set of all natural numbers and ‘o’ is operation defined on N such that aob = a^b  ∀ a,b 𝟄 N.

 Closure property :

    Let a,b 𝟄 N.

   Now aob = a^b

                   = a x a x … x b times

                   = a natural number

                   𝟄 N

  ∴ aob 𝟄 N   ∀  a,b 𝟄 N 

     N is closed under the binary operation ‘o’.

  Checking for associative law :

 Let a,b,c 𝟄 N

 Now ao(boc) = aob^c     | since boc = b^c |

                         = a^ b^c

 Also (aob) oc = a^b oc

                         = (a^b)^c

                          = a^bc

Here ao(boc) ≠ (aob)oc

 Hence N is not associative under the binary operation ‘o’

Checking for commutative :

 Let a,b,c 𝟄 N

  Now aob = a^b ≠ b^a = boa

   ∴ aob ≠ boa for all a,b

Hence N is not commutative under the binary operation ‘o’

   *** Hence Solved ***

* What is groupoid

 * What is monoid

 * What is abelian group

 * Uniqueness of identity element

 * uniqueness of  inverse element in a group

 * Is aob = a is associative in S?

#group #theory #binary #operation #abstract #algebra #syllabus #semi #groupoid #monoid #abelian #commutative

#group #theory #binary #operation #abstract #algebra #syllabus #semi #groupoid #monoid #abelian #commutative